If f : R → R be defined by f(x) = (2 – x5)1/5,

Question:

If $f: R \rightarrow R$ be defined by $f(x)=\left(2-x^{5}\right)^{1 / 5}$, then $f \circ f(x)=$

Solution:

Given: $f(x)=\left(2-x^{5}\right)^{1 / 5}$

$f o f(x)=f(f(x))$

$=f\left(\left(2-x^{5}\right)^{1 / 6}\right)$

$=\left[2-\left(\left(2-x^{5}\right)^{1 / 5}\right)^{5}\right]^{1 / 5}$

$=\left[2-\left(2-x^{5}\right)^{\frac{1}{5} \times 5}\right]^{1 / 5}$

$=\left[2-2+x^{5}\right]^{1 / 5}$

$=\left[x^{5}\right]^{1 / 5}$

$=x^{5 \times 1 / 5}$

$=x$

Hence, if $f: R \rightarrow R$ be defined by $f(x)=\left(2-x^{5}\right)^{1 / 5}$, then $f o f(x)=\underline{x} .$

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